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Math Help - Mods Help

  1. #1
    Super Member
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    Mods Help

    Hey guys, havin trouble with this question

    Using Euclids Algorithm, to find integers x and y satisfying 1127 x + 2568 y = 1, find solutions of

    1. 1127 c = 127 (mod 3568)

    Note: From using Euclids Algorithm, i got the equation 1 = 1431 \times 1127 - 452 \times 3568..

    Any help much appreciated..
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by jvignacio View Post
    Hey guys, havin trouble with this question

    Using Euclids Algorithm, to find integers x and y satisfying 1127 x + 2568 y = 1, find solutions of

    1. 1127 c = 127 (mod 3568)

    Note: From using Euclids Algorithm, i got the equation 1 = 1431 \times 1127 - 452 \times 3568..

    Any help much appreciated..
    This means that 1431 \times 1127 \cong 1 \text{ mod } 3568. What can you multiply 1431 \times 1127 by to get 127? What does this mean c is?
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    This means that 1431 \times 1127 \cong 1 \text{ mod } 3568. What can you multiply 1431 \times 1127 by to get 127? What does this mean c is?
    So its like

    1127 \times c \cong 127 \text{ mod } 3568

    which means 3568 + 127 = 3695, so theres a NUMBER (c) that 1127 multiplies by that equals 3695?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by jvignacio View Post
    So its like

    1127 \times c \cong 127 \text{ mod } 3568

    which means 3568 + 127 = 3695, so theres a NUMBER (c) that 1127 multiplies by that equals 3695?
    I'm a tad confused about what you are saying, although I suspect you are a tad confused about what a \equiv b \text{ mod }n means. If n \mid (a-b) then we write that a \equiv b \text{ mod }n. It is division, not equality. This is, essentially, just notation. So here you have that 1127 \times c \equiv 127 \text{ mod } 3568, which means that 3568 \mid (1127 \times c - 127) but this is very inconvenient to work with, so it is better to stick to the modulo notation.

    Modulo arithmetic is nice. Everything you want to work works. What we want to use here is that if a \equiv b \text{ mod }n then ac \equiv bc \text{ mod }n (This holds as a \equiv b \text{ mod }n \Rightarrow n \mid a-b \Rightarrow n \mid c(a-b) \Rightarrow ac \equiv bc \text{ mod }n). For example, 2 \equiv 7 \text{ mod }5 and so 6 \equiv 21 \text{ mod }5 (multiplying by 3).
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